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a.
Đặt \(\left\{{}\begin{matrix}\sqrt[3]{x+2}=a\\\sqrt[3]{x-2}=b\end{matrix}\right.\) ta được:
\(2a^2-b^2=ab\)
\(\Leftrightarrow\left(a-b\right)\left(2a+b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=b\\2a=-b\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}a^3=b^3\\8a^3=-b^3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=x-2\left(vô-nghiệm\right)\\8\left(x+2\right)=-\left(x-2\right)\end{matrix}\right.\)
\(\Leftrightarrow x=-\dfrac{14}{9}\)
b.
Đặt \(\left\{{}\begin{matrix}\sqrt[3]{65+x}=a\\\sqrt[3]{65-x}=b\end{matrix}\right.\)
\(\Rightarrow a^2+4b^2=5ab\)
\(\Leftrightarrow\left(a-b\right)\left(a-4b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=b\\a=4b\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}a^3=b^3\\a^3=64b^3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}65+x=65-x\\65+x=64\left(65-x\right)\end{matrix}\right.\)
\(\Leftrightarrow...\)
a) Đặt \(\sqrt[3]{65+x}=a;\sqrt[3]{65-x}=b\)
Nhận xét x = 65 không phải là nghiệm. Xét x khác 65 thì \(b\ne0\)
PT \(\Leftrightarrow a^2+b^2-5ab=0\)
\(\Leftrightarrow\left(\frac{a}{b}\right)^2-5\left(\frac{a}{b}\right)+1=0\Leftrightarrow t^2-5t+1=0\left(\text{đặt }t=\frac{a}{b}\right)\)
Hình như chị ghi đề sai, số quá xấu:((
a/ Nghiệm xấu quá
Đặt \(\left\{{}\begin{matrix}\sqrt[3]{65+x}=a\\\sqrt[3]{65-x}=b\end{matrix}\right.\) ta được:
\(a^2+b^2=5ab\Leftrightarrow a^2-5ab+b^2=0\)
\(\Leftrightarrow\left(a-\frac{5+\sqrt{21}}{2}b\right)\left(a-\frac{5-\sqrt{21}}{2}b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=\frac{5+\sqrt{21}}{2}b\\a=\frac{5-\sqrt{21}}{2}b\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\sqrt[3]{65+x}=\frac{5+\sqrt{21}}{2}\sqrt[3]{65-x}\\\sqrt[3]{65+x}=\frac{5-\sqrt{21}}{2}\sqrt[3]{65-x}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}65+x=\left(\frac{5+\sqrt{21}}{2}\right)^3\left(65-x\right)\\65+x=\left(\frac{5-\sqrt{21}}{2}\right)^3\left(65-x\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(56+12\sqrt{21}\right)x=65\left(54+12\sqrt{21}\right)\\\left(56-12\sqrt{21}\right)x=65\left(54-12\sqrt{21}\right)\end{matrix}\right.\) \(\Rightarrow x=...\)
b/ \(\Leftrightarrow\sqrt[3]{x-5}+\sqrt[3]{2x-1}=\sqrt[3]{3x+2}-2\)
\(\Leftrightarrow3x-6+3\sqrt[3]{\left(x-5\right)\left(2x-1\right)}\left(\sqrt[3]{3x+2}-2\right)=3x-6-6\sqrt[3]{3x+2}\left(\sqrt[3]{3x+2}-2\right)\)
\(\Leftrightarrow\sqrt[3]{\left(x-5\right)\left(2x-1\right)}\left(\sqrt[3]{3x+2}-2\right)=-2\sqrt[3]{3x+2}\left(\sqrt[3]{3x+2}-2\right)\)
\(\Leftrightarrow\left(\sqrt[3]{3x+2}-2\right)\left(\sqrt[3]{\left(x-5\right)\left(2x-1\right)}+2\sqrt[3]{3x+2}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+2=8\Rightarrow x=2\\\left(x-5\right)\left(2x-1\right)=-8\left(3x+2\right)\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow2x^2-13x+21=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-3\\x=-\frac{7}{2}\end{matrix}\right.\)
b, ĐKXĐ: \(x\ge\frac{5}{2}\)
\(pt\Leftrightarrow\sqrt{2x+4-6\sqrt{2x-5}}+\sqrt{2x-4+2\sqrt{2x-5}}=4\)
\(\Leftrightarrow\sqrt{\left(\sqrt{2x-5}-3\right)^2}+\sqrt{\left(\sqrt{2x-5}+1\right)^2}=4\)
\(\Leftrightarrow\sqrt{2x-5}=3\)
\(\Leftrightarrow x=7\left(tm\right)\)
a, ĐKXĐ: \(x\ge5\)
\(pt\Leftrightarrow\sqrt{x-5+4\sqrt{x-5}+4}+\sqrt{x-5+8\sqrt{x-5}+16}=0\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-5}+2\right)^2}+\sqrt{\left(\sqrt{x-5}+4\right)^2}=0\)
\(\Leftrightarrow2\sqrt{x-5}+6=0\)
\(\Leftrightarrow\sqrt{x-5}=-3\)
Phương trình vô nghiệm
Dat \(a=\sqrt[3]{65+x},b=\sqrt[3]{65-x}\)
Bien doi PT thanh \(a^2+4b^2=5ab\)
\(\Leftrightarrow a^2-5ab+4b^2=0\)
\(\Leftrightarrow\left(a^2-ab\right)-\left(4ab-4b^2\right)=0\)
\(\Leftrightarrow a\left(a-b\right)-4b\left(a-b\right)=0\)
\(\Leftrightarrow\left(a-b\right)\left(a-4b\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}a=b\left(1\right)\\a=4b\left(2\right)\end{cases}}\)
\(\left(1\right)\Leftrightarrow\sqrt[3]{65+x}=\sqrt[3]{65-x}\)
\(\Leftrightarrow65+x=65-x\)
\(\Leftrightarrow x=0\left(n\right)\)
\(\left(2\right)\Leftrightarrow\sqrt[3]{65+x}=4\sqrt[3]{65-x}\)
\(\Leftrightarrow65+x=64.65-64x\)
\(\Leftrightarrow65x=64.65-65\)
\(\Leftrightarrow x=63\left(n\right)\)
Vay nghiem cua PT la \(x=0,x=63\)
Lập phương 2 vế ta đc
\(\left(65+x\right)^2+64\left(65-x\right)^2+3\sqrt[3]{64\left(65-x\right)^2\left(65+x\right)^x}.\left(\sqrt[3]{\left(65+x\right)^2}+\sqrt[3]{\left(65-x\right)^2}\right)=125\left(65^2-x^2\right)\)
<=>\(65x^2-8190x+274625+3\sqrt[3]{64\left(65^2-x^2\right)}.\sqrt[3]{65^2-x^2}=125\left(65^2-x^2\right)\)\(65x^2-8190x+274625+3.4.\sqrt[3]{65^2-x^2}=125\left(65^2-x^2\right)\)
Đặt
\(\sqrt[3]{\left(65+x\right)}=a;\sqrt[3]{65-x}=b\) => \(a^3+b^3=130\) ta có Hpt :
\(a^2+4b^2=5ab\) (1)
\(a^3+b^3=130\) (2)
từ pt (1) => a = b Hoặc a = 4b
Thay vào pt (2) tìm ra b => a
Đặt \(a=\sqrt[3]{65+x},b=\sqrt[3]{65-x}\) thì phương trình viết thành
\(a^2+4b^2=5ab\Leftrightarrow\left(a-b\right)\left(a-4b\right)=0.\)
Suy ra \(a=b\) hoặc \(a=4b\)
Trường hợp 1. Nếu \(a=b\Leftrightarrow x=0.\) Khi đó \(A=5\cdot\sqrt[3]{65^2}\)
Trường hợp 2. Nếu \(a=4b\Leftrightarrow65+x=65\left(65-x\right)\Leftrightarrow66x=65\cdot64\Leftrightarrow x=\frac{65\cdot64}{66}\) Khi đó \(A=5\cdot65\sqrt[3]{\frac{4}{66^2}}\)
Tham khảo:
1) Giải phương trình : \(11\sqrt{5-x}+8\sqrt{2x-1}=24+3\sqrt{\left(5-x\right)\left(2x-1\right)}\) - Hoc24
Câu 4:
Giả sử điều cần chứng minh là đúng
\(\Rightarrow x=y\), thay vào điều kiện ở đề bài, ta được:
\(\sqrt{x+2014}+\sqrt{2015-x}-\sqrt{2014-x}=\sqrt{x+2014}+\sqrt{2015-x}-\sqrt{2014-x}\) (luôn đúng)
Vậy điều cần chứng minh là đúng
2) \(\sqrt{x^2-5x+4}+2\sqrt{x+5}=2\sqrt{x-4}+\sqrt{x^2+4x-5}\)
⇔ \(\sqrt{\left(x-4\right)\left(x-1\right)}-2\sqrt{x-4}+2\sqrt{x+5}-\sqrt{\left(x+5\right)\left(x-1\right)}=0\)
⇔ \(\sqrt{x-4}.\left(\sqrt{x-1}-2\right)-\sqrt{x+5}\left(\sqrt{x-1}-2\right)=0\)
⇔ \(\left(\sqrt{x-4}-\sqrt{x+5}\right)\left(\sqrt{x-1}-2\right)=0\)
⇔ \(\left[{}\begin{matrix}\sqrt{x-4}-\sqrt{x+5}=0\\\sqrt{x-1}-2=0\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}\sqrt{x-4}=\sqrt{x+5}\\\sqrt{x-1}=2\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}x\in\varnothing\\x=5\end{matrix}\right.\)
⇔ x = 5
Vậy S = {5}
Bài 1:
Đặt \(\hept{\begin{cases}S=x+y\\P=xy\end{cases}}\) hpt thành:
\(\hept{\begin{cases}S^2-P=3\\S+P=9\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}S^2-P=3\\S=9-P\end{cases}}\Leftrightarrow\left(9-P\right)^2-P=3\)
\(\Leftrightarrow\orbr{\begin{cases}P=6\Rightarrow S=3\\P=13\Rightarrow S=-4\end{cases}}\).Thay 2 trường hợp S và P vào ta tìm dc
\(\hept{\begin{cases}x=3\\y=0\end{cases}}\)và\(\hept{\begin{cases}x=0\\y=3\end{cases}}\)
Câu 3: ĐK: \(x\ge0\)
Ta thấy \(x-\sqrt{x-1}=0\Rightarrow x=\sqrt{x-1}\Rightarrow x^2-x+1=0\) (Vô lý), vì thế \(x-\sqrt{x-1}\ne0.\)
Khi đó \(pt\Leftrightarrow\frac{3\left[x^2-\left(x-1\right)\right]}{x+\sqrt{x-1}}=x+\sqrt{x-1}\Rightarrow3\left(x-\sqrt{x-1}\right)=x+\sqrt{x-1}\)
\(\Rightarrow2x-4\sqrt{x-1}=0\)
Đặt \(\sqrt{x-1}=t\Rightarrow x=t^2+1\Rightarrow2\left(t^2+1\right)-4t=0\Rightarrow t=1\Rightarrow x=2\left(tm\right)\)